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- #1

So \(\mathbf{F}_s = \mathbf{F}_r\) by Newton's 3rd. By Newton's 2nd,

\begin{align}

\mathbf{F}_s &= \mathbf{F}_r\\

\frac{d(mV_s)}{dt} &= \frac{d(mV_r)}{dt}\\

m\frac{dV_s}{dt} + V_s\frac{dm}{dt} &= m\frac{dV_r}{dt} + V_r\frac{dm}{dt}\\

V_s\cdot C &= ma_r

\end{align}

Here we assumed the steam isn't accelerating and \(\frac{dm}{dt} = 0\) on the RHS. I understand assuming the steam acceleration is zero, but why is the \(\frac{dm}{dt}\) on the RHS zero and not \(C\)?